\(n_{O_2}=\dfrac{4.8}{32}=0,15\left(mol\right)\\ Đặt:n_K=u\left(mol\right);n_{Ba}=s\left(mol\right)\left(u,s>0\right)\\ 2K+2H_2O\rightarrow2KOH+H_2\\ Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}39u+137s=17,6\\0,5u+s=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=0,1\\s=0,1\end{matrix}\right.\\ \Rightarrow\%m_K=\dfrac{0,1.39}{17,6}.100\approx22,159\%\\ \Rightarrow\%m_{Ba}\approx77,841\%\)